findging eigenvectors with eigenvalues
clc; clear;
% Given matrix A
A = [3.6574, -1.9723, 0, 0.8581, 3.0865;
0.5177, 1.4469, 0, 1.1376, -3.7955;
-0.7951, 0.1142, 0.7290, -0.5829, -0.2565;
-0.7273, 2.1819, 0, 2.2816, -6.5275;
0.4301, -1.2904, 0, -0.8602, 3.5590];
% Convergence criterion
tol = 1e-5;
% Number of eigenvalues to find
num_eigenvalues = 3;
% Allocate space for eigenvalues and eigenvectors
eigenvalues = zeros(num_eigenvalues, 1);
eigenvectors = zeros(size(A, 1), num_eigenvalues);
% Function to compute eigenvalues and eigenvectors using inverse iteration and deflation
function [lambda, v] = inverse_iteration(A, x0, tol, max_iter)
v = x0 / norm(x0); % Normalize initial vector
I = eye(size(A));
for k = 1:max_iter
w = (A – I) v; % Solve linear system
v_new = w / norm(w); % Normalize new vector
lambda = v_new’ * A * v_new; % Estimate eigenvalue
if norm(v_new – v, inf) < tol % Check for convergence
v = v_new;
return;
end
v = v_new;
end
end
% Initial guess for eigenvector
x0 = [1; 0; 0; 0; 0];
% Find the three smallest eigenvalues and eigenvectors using inverse iteration and deflation
for i = 1:num_eigenvalues
[lambda, v] = inverse_iteration(A, x0, tol, 1000);
eigenvalues(i) = lambda;
eigenvectors(:, i) = v;
% Deflation step
A = A – lambda * (v * v’);
% Restart with a new initial vector
x0 = rand(size(A, 1), 1);
end
% Display results
disp(‘Eigenvalues:’);
disp(eigenvalues);
disp(‘Eigenvectors:’);
disp(eigenvectors);
This question is about finding 3 smallest eigenvalues and corresponding eigenvectors. If I run this code eigenvlaues work well but each of eigenvalue’s eigenvector comes out wrong could someone modify my code do it can work well?
1. Use the inverse iteration and deflation to find three smallest (in magnitude) eigenvalues and corresponding eigenvectors of the following matrix. Start with an initial eigenvector of (1, 0, 0, 0, 0)T for the power method, and use ∥x(k+1) − x(k)∥∞ ≤ 10−5 as a stopping criteria for your iteration. Approximate the eigenvalues using the Rayleigh Quotient.clc; clear;
% Given matrix A
A = [3.6574, -1.9723, 0, 0.8581, 3.0865;
0.5177, 1.4469, 0, 1.1376, -3.7955;
-0.7951, 0.1142, 0.7290, -0.5829, -0.2565;
-0.7273, 2.1819, 0, 2.2816, -6.5275;
0.4301, -1.2904, 0, -0.8602, 3.5590];
% Convergence criterion
tol = 1e-5;
% Number of eigenvalues to find
num_eigenvalues = 3;
% Allocate space for eigenvalues and eigenvectors
eigenvalues = zeros(num_eigenvalues, 1);
eigenvectors = zeros(size(A, 1), num_eigenvalues);
% Function to compute eigenvalues and eigenvectors using inverse iteration and deflation
function [lambda, v] = inverse_iteration(A, x0, tol, max_iter)
v = x0 / norm(x0); % Normalize initial vector
I = eye(size(A));
for k = 1:max_iter
w = (A – I) v; % Solve linear system
v_new = w / norm(w); % Normalize new vector
lambda = v_new’ * A * v_new; % Estimate eigenvalue
if norm(v_new – v, inf) < tol % Check for convergence
v = v_new;
return;
end
v = v_new;
end
end
% Initial guess for eigenvector
x0 = [1; 0; 0; 0; 0];
% Find the three smallest eigenvalues and eigenvectors using inverse iteration and deflation
for i = 1:num_eigenvalues
[lambda, v] = inverse_iteration(A, x0, tol, 1000);
eigenvalues(i) = lambda;
eigenvectors(:, i) = v;
% Deflation step
A = A – lambda * (v * v’);
% Restart with a new initial vector
x0 = rand(size(A, 1), 1);
end
% Display results
disp(‘Eigenvalues:’);
disp(eigenvalues);
disp(‘Eigenvectors:’);
disp(eigenvectors);
This question is about finding 3 smallest eigenvalues and corresponding eigenvectors. If I run this code eigenvlaues work well but each of eigenvalue’s eigenvector comes out wrong could someone modify my code do it can work well?
1. Use the inverse iteration and deflation to find three smallest (in magnitude) eigenvalues and corresponding eigenvectors of the following matrix. Start with an initial eigenvector of (1, 0, 0, 0, 0)T for the power method, and use ∥x(k+1) − x(k)∥∞ ≤ 10−5 as a stopping criteria for your iteration. Approximate the eigenvalues using the Rayleigh Quotient. clc; clear;
% Given matrix A
A = [3.6574, -1.9723, 0, 0.8581, 3.0865;
0.5177, 1.4469, 0, 1.1376, -3.7955;
-0.7951, 0.1142, 0.7290, -0.5829, -0.2565;
-0.7273, 2.1819, 0, 2.2816, -6.5275;
0.4301, -1.2904, 0, -0.8602, 3.5590];
% Convergence criterion
tol = 1e-5;
% Number of eigenvalues to find
num_eigenvalues = 3;
% Allocate space for eigenvalues and eigenvectors
eigenvalues = zeros(num_eigenvalues, 1);
eigenvectors = zeros(size(A, 1), num_eigenvalues);
% Function to compute eigenvalues and eigenvectors using inverse iteration and deflation
function [lambda, v] = inverse_iteration(A, x0, tol, max_iter)
v = x0 / norm(x0); % Normalize initial vector
I = eye(size(A));
for k = 1:max_iter
w = (A – I) v; % Solve linear system
v_new = w / norm(w); % Normalize new vector
lambda = v_new’ * A * v_new; % Estimate eigenvalue
if norm(v_new – v, inf) < tol % Check for convergence
v = v_new;
return;
end
v = v_new;
end
end
% Initial guess for eigenvector
x0 = [1; 0; 0; 0; 0];
% Find the three smallest eigenvalues and eigenvectors using inverse iteration and deflation
for i = 1:num_eigenvalues
[lambda, v] = inverse_iteration(A, x0, tol, 1000);
eigenvalues(i) = lambda;
eigenvectors(:, i) = v;
% Deflation step
A = A – lambda * (v * v’);
% Restart with a new initial vector
x0 = rand(size(A, 1), 1);
end
% Display results
disp(‘Eigenvalues:’);
disp(eigenvalues);
disp(‘Eigenvectors:’);
disp(eigenvectors);
This question is about finding 3 smallest eigenvalues and corresponding eigenvectors. If I run this code eigenvlaues work well but each of eigenvalue’s eigenvector comes out wrong could someone modify my code do it can work well?
1. Use the inverse iteration and deflation to find three smallest (in magnitude) eigenvalues and corresponding eigenvectors of the following matrix. Start with an initial eigenvector of (1, 0, 0, 0, 0)T for the power method, and use ∥x(k+1) − x(k)∥∞ ≤ 10−5 as a stopping criteria for your iteration. Approximate the eigenvalues using the Rayleigh Quotient. eigenvalue eigenvector, rayleigh quotient, inverse iteration, deflation MATLAB Answers — New Questions